Such big numbers!

Algebra Level pending

Let p be a prime number where p cubed is equal to the sum of 101 consecutive positive integers.

Find the smallest of the 101 consecutive numbers.

1 solution

Damiann Mangan
Mar 14, 2014

We could easily count the sum $101$ consecutive number as, where the smallest is $n+1$ ,

$\frac{(n+101)(n+102)}{2}-\frac{n(n+1)}{2}=101(n+51)$

which means it have $101$ as a factor. This could only means that the prime number is actually $101$ .

$n = \frac{101^{3}}{101}-51=10150$

Therefore, the smallest of the consecutive numbers is $n+1=10151$ .

i didnt understand :/

Saqlain Shoaib - 7 years, 2 months ago